Problem 10711
(American Mathematical Monthly, Vol.106, February 1999) Proposed by F. Luca (Germany).
A natural number isperfect if it is the sum of its proper divisors. Prove that two consecutive num- bers cannot both be perfect.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Firenze, viale Morgagni 67/A, 50134 Firenze, Italy.
We denote by σ(n) the sum of the divisors of the natural number n. It is easy to see that the arithmetic function σ is multiplicative that is
σ(a · b) = σ(a) · σ(b) iff (a, b) = 1.
Then n is a perfect number if and only if σ(n) = 2n. Moreover the following two elementary properties of perfect numbers hold:
(i) if n is even then n is perfect if and only if
n= 2q−1(2q−1) where q and 2q−1 are primes.
(ii) if n is an odd perfect number then it must have the form n= p4m+1s2 where p is a prime such that [p]4= 1 and (p, s) = 1.
For a proof of the above results see for example L. Dickson, History of the theory of numbers, vol.
1, 1919, pp. 1-33. Note that the question as to whether there exist odd perfect numbers is as yet unsolved.
Therefore, by (i), two consecutive numbers are both perfect if
n−= 2q−1(2q−1) − 1 or n+= 2q−1(2q−1) + 1
is perfect. The prime q is not 2 because neither n−= 5 nor n+= 7 is perfect. Hence q = 2a + 1 ≥ 3.
1) Assume that n− is perfect. Then
[n−]4= [4a]4·[2 · 4a−1]4+ [−1]4= [3]4.
By (ii), n− = p4m+1s2 with [p]4 = 1. Since s is odd then either [s]4 = 1 or [s]4 = 3 and in both cases [s2]4= 1. So we have the contradiction
[3]4= [n−]4= [p]44m+1·[s2]4= [1]4. 2) Assume that n+ is perfect. Then
[n+]3= [4a]3·[2 · 4a−1]3+ [1]3= [1]3+ [1]3= [2]3.
By (ii), n+= p4m+1s2 where p is a prime such that (p, s) = 1. Now, it easy to verify that [p4m+1]3·[s2]3= [2]3 iff [p]3= [2]3 and [s2]3= 1.
Moreover, since p is prime,
[σ(p4m+1)]3= [1 + p + . . . + p4m+ p4m+1]3= [1 + 2 + . . . + 1 + 2]3= [0]3. By the multiplicative property of σ, we have the contradiction
[1]3= [2n+]3= [σ(n+)]3= [σ(p4m+1)]3·[σ(s2)]3= [0]3.